In the correctly worked addition problem above, M, N, R, S, T and V

VERITAS PREP OFFICIAL SOLUTION:

Solution: This is certainly harder than the PS question but our process will remain the same.

First, let’s see what information we are given in the question – the units digits of all three numbers are the same. The two-digit numbers add up to give a three digit number. The hundreds digit, S, is either 1 or 2. Three two-digit numbers cannot add up to give a number 300 or more since 99 + 99 + 99 = 297. We have no information on what the value of R can be. All we know is that R cannot be 0 because 0+0+0 = 0 but V needs to be different from R.

Let’s look at the statements now.

Statement 1: M, N and P are positive even integers.

At first, it may seem that this has nothing to do with the value of R but we must analyze what is given to be sure.

M, N and P must take distinct values out of 2, 4, 6 and 8 and add up to give the units digit of T (again, distinct)

Every time you add three even numbers, you will get an even number. Let’s see which combinations we can get:

2 + 4 + 6 = 12

2 + 4 + 8 = 14

2 + 6 + 8 = 16

4 + 6 + 8 = 18

Note that in all four cases, the units digit is one of the numbers but T must be distinct. This means that there must have been a carryover from the previous addition. So when we added the three Rs, we must have got a carryover. Had R been 3 or less, we would not have got a carryover since 1+1+1 = 3, 2+2+2 = 6 and 3+3+3 = 9. So R must be greater than 3.

One such case would be

This statement alone is sufficient.

Statement 2: S = 2

The result of addition gives us a number which is more than 200. In statement 1 we saw a case in which S is 2 and R is greater than 3. Now all we have to do is find a case in which S is 2 and R is less than 3. One of these cases is

So this statement alone is not sufficient.

Answer (A)